easy with projetive geometry. let $CD\cap{AB} = M$. we know that $P,C,M,D$ are one harmonic quadruple, so, $AP,AE,AB,AF$ are one harmonic beam. $AP||EF$ implies that $B$ is the midpoint of $EF$.

Since the tangets to the circumscribed circle of $ACBD$ intersect on the diagonal of it we conclude that $ACBD$ is a harmonic quadrilateral. But then $AB$ is the symedian of $\triangle{ACD}$. Moreover $\angle{BEA}=\angle{EAP}=\angle{PDA}=\angle{CDA}$ and hence $CD$ and $EF$ are antiparallel with respect to $\angle{EAF}$. Consequently a symedian in $\triangle{ACD}$ is a median in $\triangle{AEF}$. Q.E.D.

Remark. $N\in (CD),\ NC=ND;\ S\in AN\cap \overline {EBF}\Longrightarrow$ the point $S$ belongs to the given circle.

I have a little bit long proof using Euclidean Geometry. $pf$) Let $T \in circumcircle$ s.t $BT \parallel PD$, $S=AT \bot CD$. Here, $\angle PBA = \angle BTA = \angle TSD = \angle ASC$, or $ASBP$ is cyclic. And cuz $ABPO$ is cyclic, we get $ABPOS$ is cyclic. ($O$ is the circumcenter) therefore, $\angle OSP = \angle OAP = \frac{\pi}{2}$. or $DS = CS$. and cuz $CD \parallel BT$, $BC = DT$. or $\angle BAC = \angle TAD$......$(*)$. And, $\angle FEA = \angle CAP = \angle CDA$ or $\triangle AEF$ resembles $\triangle ADC$. cuz $AS$ is median of $\triangle ADC$, we get $AB$ is the median of $\triangle AEF$ from $(*)$. therefore, we get $BE=BF$. $Q.E.D.$

the problem is fromchina(westen mathematical olympiad)chendu,sichuan $2005$.geometry of first day. problem $2$ . ok lets solve the problem: suppose $(C)$ is a circle and $P$ is a point in the exterior of it.each time we draw a variable line and let it. intersect the circle in points $C,D$. the locus of points like $K$,inwhich we have:$(PKCD)=-1$, is a line which we name it polar. name the intersection $PD$ and $AB,K$. its clear that $(PKCF)=-1$ theorem:we have,$(ABCD)=-1$.and $O$ is a arbitrary point in plane. for each line which intersects $PA,PB,PC,PD$, in points $A',B',C',D'$,we have:$(A'B'C'D')=-1$ the we have:$A(PKCF)=-1$ then the points $F,B,E$,and teh intersection of $AP$ and $FE$ are harmonic. but $AP$ and $EF$ are paralell. then: $FB=BE$

Easy to get that : $ {D,K,C,P}$ are harmonic points group ;draw parallel line to $ {PD}$ through $ {B}$ $ \Longrightarrow$ $ \frac{PC}{CK}=\frac{PD}{DK}$$ \Longrightarrow$ $ {AT}\frac{YB}{YT}={AT}\frac{XB}{XT}$;By similar triangles : we get : $ {BE}={BF}$...$ {Q}$.$ {E}$.$ {D}$

Attachments:

First of all we note that the quad $ACBD$ is harmonic,thus meaning that $\frac{AC}{BC} = \frac{AD}{BD}$ So now first of all note that $\triangle CAB \sim \triangle BAE$.This is because $\angle BEA = \angle PAC = \angle CDA = \angle CBA$ and becaus e $\angle BAC = \angle EAB$,Thus from here we get that $EB= \frac{CB}{AB} EA$.From the power of point we have that $EB.EF=EC.EA$,thus we have the following: $$\frac{CB}{AB}.EF = EC \iff \frac{CB}{AB} = \frac{CE}{EF}$$With the conditions of a same angle $\angle CEF = \angle CBA $,we have that $\triangle CEF \sim \triangle CBA $, which implies $\angle ECF = \angle ACB$. Now we do a simple angle-chase: $$\angle ECF = \angle ACB \implies \angle ECB +\angle BCF = \angle ACF + \angle FCB \implies \angle ECB = \angle ACF$$Because the quad $ACBD$ is cyclic we have that $\angle ECB = \angle BDA = \angle BAP = \angle ABF$,thus we have $\triangle ABF \sim \triangle ADB$. Now from this similarity we get that $BF = \frac{AB}{AD}.BD$.So now let's see if that equals $\frac{CB}{AB}.EA$ $$\frac{AB}{AD} BD = \frac{CB}{AB} EA \implies \frac{AB}{AD} \frac{BD}{AD} = \frac{CB}{AB} \implies \frac{AC}{AB} \frac{BD}{AD} = \frac{CB}{AB}$$Now this implies that $\frac{BD}{AD} = \frac{BC}{AC} $ Which is true,because we have that the quad $ACBD$ is harmonic,which means $(A,B;C,D)=-1$,and the ratio directly follows from this. Thus $BE=BF$.....